# Tag Info

18

The identity $$V = K \frac{dV}{dt}$$ is only guaranteed with a constant $K$ if your assumptions actually hold. The first identity $V=RI$ only holds for a resistor, while the other holds for a capacitor. So in this sense, the letters $V,I$ in these equations mean something else. In one of them, it's the current through (or voltage on) a particular resistor, ...

14

For static charges, the relationship is V (voltage) = Q (charge) / C (capacitance). Capacitance is a function of the shape, size and distance between objects, which are all continuous values. (Well, I suppose you could argue that shape and size are quantized to the atomic spacing of the object's material, but you can't say the same thing for distance.) So ...

7

Voltage is a continuous function. If you are a certain distance from a (point) charge $q$, the potential is $$V=\frac{q}{4\pi\epsilon_0 r}$$ By adjusting the value of $r$ to anything you want (not quantized), you can get any potential you want. And so yes, when you do any analog-to-digital conversion, you will "destroy" a certain amount of information. ...

6

This is true but strictly limited to RC circuits without external sources: that is, a resistor hooked up to a capacitor with nothing else in between. In that case, $V$ is indeed proportional to $\dot V$, with a crucial minus sign in between: $$\dot V=-\frac1\tau V,$$ where $\tau>0$ is some constant. This equation implies that $V(t)=V(0)e^{-t/\tau}$, ...

5

Voltage doesn't come directly from the charge of the electron. It's the energy per charge. The charge carriers may be discrete, but the energy is not. We can easily generate a potential by moving a wire through a magnetic field. The potential is proportional to the speed of the wire, which is a continuous value. $$V = vBL\sin{\theta}$$

5

Generally, no. You've written two equations. The first relates voltage and current for an isolated resistor. The second relates voltage and current for an isolated capacitor. Given only those two expressions, there's no reason at all that they can or should be combined. That is, you have no circuit, only isolated components. There are principles for ...

4

It's not a fundamental feature of electrical potential, but: If you have a polycrystalline metal and you cut and polish a smooth surface, the differently-oriented regions will present a different lattice plane to the outside. Crystals cut along different planes may have slightly different work functions, and so the electric potential very close to such a ...

3

As Kevin Reid aptly explains, the circuit you have drawn is not realizable. But, let's take the closest physical thing you could build, assuming: your voltage source can supply enough energy that we don't hit its limits like all physical things, this apparatus has non-zero size Then, the circuit you actually built is this: simulate this circuit ...

3

It is quite easy to use a step-up converter to generate almost any voltage. The question is "voltage at what current". The power a battery can deliver is finite (power = voltage times current), but you can convert voltages in many different ways. The most obvious is an oscillator (inverter) followed by a transformer and a rectifier, but there are more ...

3

The $KV$ in $30KV$ refers to the motor velocity constant $K_v$. It does not refer to kilovolt $kV$, cf. comments by John Rennie.

3

The simple reason is that you have only passive, linear components in your circuit; when you push with a certain voltage, you get a certain current response. If the voltage cycles a certain number of times per second, that's how often the current must respond. Think about it the other way - if you _didn't_have the same frequency, what response would make ...

2

The 3 A that leaves the battery depends on two things: the amount of supply voltage, and the impedance (here resistance) that draws this current. So the second part simply says: The 3 A current that leaves the voltage source = The voltage source divided by the total resistance (which is the internal resistance, plus the equivalent of the parallel ...

2

Kirchoff's laws tell us that the potential drop across any closed loop in a circuit must be equal to the voltage sources in the loop, from which we conclude that the voltage drop across resistors in parallel must be equal. Ohm's law states: $$V=IR$$ From which we conclude that, since $V$ is fixed, if the different resistors have different $R$'s, then the ...

2

First, the reactive power is not dissipated, but which corresponds to power delivered by the power stored in the reactive component (inductor or capacitor) during a semi-cycle; in the next half cycle, the component returns the stored energy to the source. For this to occur, the component must have the ability to store energy. In the case of a capacitor, the ...

2

Current flows in a closed loop, so $$I_c = I_r\tag1$$ There is a voltage across the capacitor due to the integrated charge, across the resistor due to the instantaneous current, and across the whole loop due to the motion of the resistor. The net voltage around the loop is zero, so $$V_c - I_r R + \mathcal{E}=0\tag2$$ The force on the wire is given by ...

2

You are massively overthinking the problem. The collector current is given (by the diagram) to be 150x the base current. The sum of base and collector current has to flow through the emitter... That's all you need to solve this. In particular, a current source will look to a circuit like "whatever resistance" it needs to be in order for the correct current ...

2

You're quite correct that there will be some current flowing, so there must be a voltage drop due to the internal resistance of the battery. The EMF measured by any voltmeter will always be less than the true EMF. If the internal resistance of the battery is $R_b$ and the resistance of your voltmeter is $R_m$ then the voltage you measure will be: $$V = ... 2 AC has each wire positive half the cycle and negative the other half. It will charge the capacitor on one half cycle and discharge it on the other half. The net charge will be zero. 2 Of course you can charge a capacitor with AC. The problem is that you keep changing how it is charged. While you apply a positive voltage to one plate, it will get a positive charge; half a cycle later, it will attempt to get a negative charge; and so it continues. The capacitor is always a little bit behind - as your AC voltage is changing, the capacitor ... 1 Like any set of equations in physics, this is true when the conditions stated apply. Here, you are using two equations:$$ V(t) = RI(t) $$when the current passes through an element that is a perfect resistor, and$$ C \frac{dV}{dt} = I(t) $$when the current passes through an element that is a perfect capacitor. So, in a simple RC circuit consisting of a ... 1 take only first 2 resistors and rest as x now the vertical resistor and your x will be in parallel, effective resistance would be Rx/R+x with series in horizontal resistor. Now equivalent resistance would be$$Req. = R+ (Rx/R+x). $$take Req. as x again Form quadratic equation and solve for x. This will be the answer. 1 You are increasing the number of electrons in the wire, but only by a very small amount. There's a somewhat clichéd but still excellent analogy for electrical circuits called the hydraulic analogy. In the hydraulic analogy the power supply is a pump, and the pressure is the voltage. The water represents the electrons, so the pressure generated by the pump ... 1 yes, you will have to pay more if your load is inductive. most of energy meters work with the voltage and current to calculate energy. When we have inductive load it takes more current than resistive load to produce same power or output. P=V.I.Cosx where x is the phase angle between voltage(V) and current(I). if the phase angle x goes higher, the power ... 1 I seriously doubt that the batteries were putting out 30 kV. You surely misread or misheard something. The chemistry of batteries is such that individual cells produce from a few 100 mV to a few volts. A 30 kV battery would require 1000s of cells, which would make no sense at all. In addition, 30 kV is much more difficult to handle and would be much less ... 1 As radio amateurs we've all learned the various relationships of power, voltage, current and resistance as expressed in Ohm's Law Ohm's law is:$$ E = IR \tag{1} $$This doesn't directly say anything about power. There is the related Joule's first law, which relates to electrical power converted to heat in resistive materials:$$ P = I^2 R \tag{2} $$... 1 The two resistors are in parallel. This means that at A the current splits between them relating to their reistance. So the current throw the top resistor is 3\,A and throw the bottom resistor is 1 \,A. If we use Kirchhoff's current law which states, that in any node (like A) the current flowing into the node is equal to the current flowing out of ... 1 This is a parallel circuit, not a series one. So the currents not need be the same, but the potencial difference is the same instaed. In a series circuit the current is the same but the potencial is different in the many elements of the circuit. You can think about this like the current is a flow of water, since is parallel, the current (the flow) "divides" ... 1 A voltage or current given as a complex constant is a phasor. A voltage given as the complex constant V_z represents the real voltage$$V(t) = \operatorname{Re} \left( V_z e^{i\omega t} \right)\ \ ,$$where \omega is the voltage's angular frequency and t is time. Currents represented as phasors work the same way. 1 perhaps because a resistor (at least an ideal resistor) is not a reactive component. and neither do reactive components (such as capacitors and ideal inductors) dissipate power. 1 Would you post an answer for me to accept? It is stipulated that the 18V voltage delivers 8A which I interpret to mean that the 8A leaves the positive terminal of the source to enter the top node thus, the currents entering the top node sum to$$8A + 13A$$The currents leaving the top node sum to$$\frac{18V}{R_A} + 3A Setting both sums equal ...

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